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ArcactCatmedeq8

Answered question

2021-11-17

Use the definition of Taylor series to find the Taylor series, centered at c for the function.

f (x) = \ln x, c = 1

Answer & Explanation

Sact1989

Beginner2021-11-18Added 10 answers

The Taylor series of $\ln x$ around c=1. We know from the definition of Taylor series of the function f
$f (x) = f (x) ∣_{x = c} + \frac{f^{'} (x) ∣_{x = c}}{1!} (x - c) + \frac{f^{″} (x) ∣_{x = c}}{2!} {(x - c)}^{2} + \frac{f^{‴} (x) ∣_{x = c}}{3!} {(x - c)}^{3} + \dots$
Now the values of differential functions at given point
$f (1) = \ln (1) = 0,$ $f^{'} (1) = \frac{1}{x} |_{x = 1} = 1$
$f^{″} (1) = - {\frac{1}{x^{2}}}_{x = 1} = - 1, f^{‴} (1) = \frac{2}{x^{3}} {|_{{x = 1}} = 2 f^{⁗} (1) = - \frac{6}{x^{4}} |}_{x = 1} = - 6$
$f^{n} (1) = {(- 1)}^{n + 1} \frac{(n - 1)!}{x^{n}} ∣_{x = 1} = {(- 1)}^{n} ((n - 1)!)$
Thus the Taylor series expression become
$f (x) = 0 + (x - 1) - \frac{1}{2!} {(x - 1)}^{2} + 2! \frac{1}{3!} {(x - 1)}^{3} - 3! \frac{1}{4!} + {(- 1)}^{n + 1} (n - 1)! \frac{1}{n!} {(x - 1)}^{n}$
Or we can write as
$f (x) = (1 - x) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3} - \frac{1}{4} {(x - 1)}^{4} + {(- 1)}^{n + 1} (n - 1)! \frac{1}{n!} {(x - 1)}^{n} - \dots$

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